The relationship between Ricci and Gaussian curvatures

Question

Why do we have that for a surface (dimension $2$) that

$$\text{Ric}(X, Y) = K \langle X, Y \rangle ,$$

where $K$ is the Gaussian curvature and $X, Y$ are vector fields?

Answer 1

Another way to see this is from the following description of the Ricci tensor evaluated on a unit length vector $X \in T_pM$: For a manifold $(M,g)$ of dimension $n$, it is given by $$\text{Ric}(X,X) = \sum_{j=2}^{n}k(\pi_{X,e_j})$$ where $(X,e_2, \ldots, e_n)$ is an orthonormal basis of $T_pM$ and where $k(\pi_{X,e_j})$ is the sectional curvature of the plane $\pi_{X,e_j}$ generated by $X$ and $e_j$ inside $T_pM$. Since the sectional curvature of a plane $\pi$ is just the Gaussian curvature at $p$ of the surface given by $\exp(\pi) \subset M$, when $n=2$ this surface is a whole open set of $M$ and we get $$ \text{Ric}(X,X) = k(\pi_{X,e_2}) = K_M(p) $$ for $K_M$ the gaussian curvature of $M$. Since $\text{Ric}$ at a point $p$ is entirely determined by its values on unit vectors as above, it is easy to deduce that you get in fact $\text{Ric} = K_Mg$ as tensors.

Answer 2

For a surface $(M, g)$, there is only independent component of the curvature tensor $R$, namely, $$R_{1212} = - R_{1221} = -R_{2112} = R_{2121},$$ and this quantity (which depends on the choice of coordinates) is related to the Gaussian curvature $K$, by $$R_{1212} = -\det(g) K$$ (which is independent of coordinates). (Beware that this sign is a matter of convention, and not everyone uses the same one.)

This leads to a decomposition of $R$--- $$R_{abcd} = K(g_{ac} g_{bd} - g_{ad} g_{bc})$$ ---and taking an appropriate trace, say, with $g^{ac}$ gives the formula you mention: $$\text{Ric}_{bd} = K g_{bd}.$$